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Chapter 3 EXAMPLE 3.1 S for a phase change
The Second Law of Thermodynamics
Find S for the melting of 5.0 g of ice (heat of fusion 79.7 cal/g) at 0°C and
1 atm. Find S for the reverse process.
The melting is reversible and Eq. (3.25) gives
¢H 179.7 cal>g215.0 g2
¢S 1.46 cal>K 6.1 J>K (3.26)
T 273 K

For the freezing of 5.0 g of liquid water at 0°C and 1 atm, q is negative, and
rev
S 6.1 J/K.
Exercise
The heat of vaporization of water at 100°C is 40.66 kJ/mol. Find S when 5.00 g
of water vapor condenses to liquid at 100°C and 1 atm. (Answer: 30.2 J/K.)

2
4. Reversible isothermal process. Here T is constant, and S T 1 dq
1 rev
2
T 1 dq q /T. Thus
1 rev rev
¢S q >T rev. isotherm. proc. (3.27)
rev
Examples include a reversible phase change (case 3 in this list) and two of the four
steps of a Carnot cycle.
5. Constant-pressure heating with no phase change. First, suppose the heating is
done reversibly. At constant pressure (provided no phase change occurs), dq
rev
2
dq C dT [Eq. (2.51)]. The relation S dq /T [Eq. (3.21)] becomes
P P 1 rev
T 2 C P
¢S dT const. P, no phase change (3.28)
T 1 T
If C is essentially constant over the temperature range, then S C ln (T /T ).
P P 2 1

EXAMPLE 3.2 S for heating at constant P

The specific heat capacity c of water is nearly constant at 1.00 cal/(g °C) in the
P
temperature range 25°C to 75°C at 1 atm (Fig. 2.15). (a) Find S when 100 g of
water is reversibly heated from 25°C to 50°C at 1 atm. (b) Without doing the cal-
culation, state whether S for heating 100 g of water from 50°C to 75°C at 1 atm
will be greater than, equal to, or less than S for the 25°C to 50°C heating.
(a) The system’s heat capacity is C mc (100 g)[1.00 cal/(g °C)]
P P
100 cal/K. (A temperature change of one degree Celsius equals a change of one
kelvin.) For the heating process, (3.28) with C constant gives
P
T 2 dq rev T 2 C P T 2
¢S dT C ln
P
T 1 T T 1 T T 1
323 K
1100 cal>K2 ln 8.06 cal>K 33.7 J>K
298 K
(b) Since C is constant, the reversible heat required for each of the
P
processes with T 25°C is the same. For the 50°C to 75°C change, each in-
finitesimal element of heat dq rev flows in at a higher temperature than for the
25°C to 50°C change. Because of the 1/T factor in dS dq /T, each dq rev
rev